   Chapter 10.4, Problem 1E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Return to sales The manufacturer of GRIPPER tires modeled its return to sales from television advertising expenditures in two regions as follows: Region  1 :   S 1 =   30   +   20 x 1 —   0.4 x 1 2 Region  2 :   S 2 =   20   +   36 x 2 −   1.3 x 2 2 where S 1  and   S 2 are the sales revenue in millions of dollars and x 1  and   x 2 are millions of dollars of expenditures for television advertising.(a) What advertising expenditures would maximize sales revenue in each district?(b) How much money will be needed to maximize sales revenue in both districts?

(a)

To determine

To calculate: The advertising expenditures required to maximize sales revenue in the two regions S1 and S2.

Explanation

Given Information:

Sales revenue S1 and S2 are

Region 1:S1=30+20x10.4x12

Region 2:S2=20+36x21.3x22

Where x1 and x2 are the advertising expenditures.

Formula used:

To find the maximum value, calculate the relevant stationary value of an equation. Differentiate the function with respect to the independent variable and equate it to 0. the relevant value is the stationary value that satisfies the provided conditions.

The power rule is used for a function in which the expression can be written as every term raised to a power (be it fractional, positive or negative). For the function f(x)=xn, the derivative is

ddx[xn]=nxn1

Calculation:

Consider the provided equations,

Sales revenue S1 and S2 are

Region 1:S1=30+20x10.4x12

Region 2:S2=20+36x21.3x22

To find the critical value for S1=30+20x10

(b)

To determine

To calculate: The money required to maximize sales revenue in the two regions.

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